Continuous-time Fourier transform: from omega to f

In ECE301, you learned that the Fourier transform of a step function $ x(t)=u(t) $ is the following:

$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ). $

Use this fact to obtain an expression for the Fourier transform $ X(f) $ (in terms of frequency in hertz) of the step function. (Your answer should agree with the one given in this table.) Justify all your steps.

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Answer 1

There are a few things that you need to know to accomplish this problem. The two main formulas that you need are $ \omega = 2 \pi f $ and $ \delta(cx)= \frac{1}{c} \delta(x) $ for c>0.

PROOF

$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $

$ y=cx => \frac{dy}{c}=dx $

$ \int_{-\infty}^\infty \delta(y)\frac{dy}{c}=\frac{1}{c} $

THEREFORE

$ \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $

and

$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ) = \frac{1}{2}(\frac{1}{j\pi f} + \delta(f)) $

-my

Instructor's comments: In essence, this answer is correct, except that you forgot to state that

$ \delta(cx)= \left\{ \begin{array}{ll} \infty, & x = 0,\\ 0, & \text{ else.} \end{array} \right. $

However the "flow of thoughts" is a bit hard to follow. I would suggest a slight reordering/rearrangement of the arguments. You could save space by giving less details regarding the change of variables when integrating. -pm

Answer 2

I claim that $ c\delta(cx)= \delta(x) $ because of the following two facts:

$ c\delta(cx)= \left\{ \begin{array}{ll} \infty, & x = 0,\\ 0, & \text{ else.} \end{array} \right. $

and

$ \int_{-\infty}^\infty c\delta(cx)dx=\int_{-\infty}^\infty c\delta(y)\frac{dy}{c} = 1 $

Now, using the equation $ \omega=2 \pi f $, we have:

$ X(f)=\frac{1}{j2\pi f} + \pi\delta(2\pi f) = \frac{1}{j2\pi f} + \frac{1}{2}\delta(f) $

Instructor's comments: Pretty good! What do you guys think? -pm

Answer 3

$ c\delta(cx)= \delta(x) $ could be proven this way.

$ \begin{align}\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t)e^{j\omega t} dt \\= X(\omega)\end{align} $

$ \begin{align}\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t)e^{j\omega t} dt \\= \int_{-\infty}^{\infty} x(t)e^{j2\pi f t} dt\end{align} $

If you set $ u = 2\pi t, du = 2\pi dt $

$ \begin{align}\int_{-\infty}^{\infty} x(t)e^{j2\pi f t} dt=\int_{-\infty}^{\infty} x(u\frac(2\pi))e^{j2\pi f t} dt \end{align} $

more to be added

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